Question

Consider the following unbalanced chemical equation for the combustion of pentane(C₅H₁₂):

${\mathit{C}}_{\mathbf{5}}{\mathit{H}}_{\mathbf{12}}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}$

If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100% yield?

Short answer

The mass of water is produced when 20.4 g of pentane are burned in excess oxygen is 30.63G.

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present in the reactant side of the equation equals the number of atoms present in the product side of the equation.

Balancing the given equation

The chemical equation given as,

$C_5{H}_{12}\left(I\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(I\right)$

This is an unbalanced equation. So, first and foremost, the equation must be balanced. In the reactants, there are 5 C atoms, however in the products, there are just 1. So, first, balance the C atoms by assigning a coefficient of 5 to CO, yielding the following equation:

$C_5{H}_{12}\left(I\right)+O_2\left(g\right)\to 5C{O}_2\left(g\right)+H_{2}O\left(I\right)$

Following that, theH atoms may be balanced by adding a coefficient 6 to, yielding the equation:

$C_5{H}_{12}\left(I\right)+O_2\left(g\right)\to 5C{O}_2\left(g\right)+6H_{2}O\left(I\right)$

The number of O atoms in the products is now16, whereas the number in the reactant is just two. Placing a coefficient of 8 on O₂ will also balance the O atoms, resulting in the balanced equation shown below.

$C_5{H}_{12}\left(I\right)+8O_2\left(g\right)\to 5C{O}_2\left(g\right)+6H_{2}O\left(I\right)$

Determination of moles of C5H12

Pentane has a molar mass of $72.15\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$. As a result, the number of moles of pentane present in 20.4g pentane is first calculated:

Moles of$\begin{array}{rcl}{C}_5{H}_{12}& =& \frac{20.4g}{72.5{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}{C}_5{H}_{12}}\\ & =& 0.2827mol{C}_5{H}_{12}\end{array}$

Determination of moles of H2

On combustion, one mole of pentane creates 6 moles of water, assuming a 100% yield, according to the balanced equation. As a result, the number of moles of water created by 0.2827molC₅H₁₂ may be computed as follows:

Moles of $\begin{array}{rcl}{H}_2& =& \left(\frac{6mol{H}_{2}O}{1mol{C}_5{H}_{12}}\right)\left(0.2827mole{C}_5{H}_{12}\right)\\ & =& 1.70mol{H}_{2}O\end{array}$

As a result, 1.70 moles of water are created by the burning of 20.4 g pentane. Because a mole of water weighs 18.02 g, the amount of water in 1.70 moles may be calculated as follows:

role="math" localid="1655215369753" $\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}& =& 1.70\mathrm{mol}\left(\frac{18.02\mathrm{g}}{1\mathrm{mole}}\right)\\ & =& 30.63\mathrm{g}\end{array}$

Thus, the mass of water generated by 20.4 g pentane combustion is 30.63 g .